41st Balkan Mathematical Olympiad

The inequality can be rewritten as $$2S\left(\frac{1}{\sin \alpha }+\frac{1}{\sin \beta }+\frac{1}{\sin \gamma }-\frac{4}{\sqrt{3}}\right)\ge P{A}^2+P{B}^2+P{C}^2.$$ Note that $AB\cdot AC\cdot \sin \alpha =AB\cdot BC\cdot \sin \beta =AC\cdot BC\cdot \sin \gamma =2S$ and $$2S=2(S_{PAB}+S_{PBC}+S_{PCA})=(PA\cdot PB+PB\cdot PC+PC\cdot PA)\frac{\sqrt{3}}{2}.$$ The inequality can be further rewritten as $$AB\cdot AC+AC\cdot BC+BC\cdot AB-2(PA\cdot PB+PB\cdot PC+PC\cdot PA)\ge P{A}^2+P{B}^2+P{C}^2$$ that is equivalent to $$AB\cdot AC+AC\cdot BC+BC\cdot AB\ge (PA+PB+PC{)}^2$$ Consider the points $X$ and $Y$ such that $△XAB$ and $△YAC$ are equilateral ($X$ and $C$ lie on different halfplanes with respect to $AB$, similarly $Y$ and $B$ with respect to $AC$).

$△AXB+△APB=180^{\circ }=△AYC+△APC\Rightarrow XAPB$ and $YAPC$ are cyclic quadrilaterals. Also note that $△BPA+△APY=120^{\circ }+△ACY=120^{\circ }+60^{\circ }=180^{\circ }$ hence $B,P$ and $Y$ are collinear. Similarly, points $C,P$ and $X$ are collinear. From Ptolemy's Theorem in cyclic quadrilateral $XAPB$ we have that $PA\cdot XB+PB\cdot XA=PX\cdot AB$, but since $△XAB$ is equilateral, then $XA=XB=AB$ and we have that $PA+PB=PX$. From here, $CX=PX+PC=PA+PB+PC$. Similarly, $BY=PA+PB+PC$.

Now, we apply Ptolemy's Inequality in quadrilateral $XBCY$ and get that $XB\cdot YC+XY\cdot BC\ge CX\cdot BY$. From Triangle Inequality we have that $AX+AY\ge XY$ so $XB\cdot YC+(AX+AY)BC\ge CX\cdot BY$. Rewriting the inequality based on the above relations we have that $AB\cdot AC+AB\cdot BC+AC\cdot BC\ge (PA+PB+PC{)}^2$.

The equality occurs if and only if both equality cases of Ptolemy's Inequality and Triangle's Inequality occur. The equality case of Triangle's Inequality occurs when $X,A$ and $Y$ are collinear $⟺180^{\circ }=△XAB+△BAC+△CAY=60^{\circ }+△BAC+60^{\circ }\Rightarrow △BAC=60^{\circ }$. The Ptolemy's Inequality equality case occurs if and only if $XBCY$ is cyclic $⟺180^{\circ }=△BXY+△BCY=60^{\circ }+△BCA+△ACY=60^{\circ }+△BCA+60^{\circ }\Rightarrow △BCA=60^{\circ }$. So the equality case happens if and only if $△ABC$ is equilateral. $\square$

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Alternative solution.

Applying cotangent rule for $△PAB$, \triangle PBC$and$\triangle PAC$wehave:$a^2 + b^2 + c^2 = 2p^2 - 2r^2 - 8rR$andweneedtoprovethat$a^2 + b^2 + c^2 \le 9R^2$andEule{r}^{\prime }sinequality$r \le \frac{R}{2}$wehave$\triangle ABC$isequilateral.$\square$